1
JEE Main 2016 (Online) 9th April Morning Slot
+4
-1
For x $$\in$$ R, x $$\ne$$ 0, Let f0(x) = $${1 \over {1 - x}}$$ and
fn+1 (x) = f0(fn(x)), n = 0, 1, 2, . . . .

Then the value of f100(3) + f1$$\left( {{2 \over 3}} \right)$$ + f2$$\left( {{3 \over 2}} \right)$$ is equal to :
A
$${8 \over 3}$$
B
$${5 \over 3}$$
C
$${4 \over 3}$$
D
$${1 \over 3}$$
2
JEE Main 2016 (Offline)
+4
-1
If $f(x)+2 f\left(\frac{1}{x}\right)=3 x, x \neq 0$, and $\mathrm{S}=\{x \in \mathbf{R}: f(x)=f(-x)\}$; then $\mathrm{S}:$
A
is an empty set.
B
contains exactly one element.
C
contains exactly two elements.
D
contains more than two elements.
3
AIEEE 2011
+4
-1
The domain of the function f(x) = $${1 \over {\sqrt {\left| x \right| - x} }}$$ is
A
$$\left( {0,\infty } \right)$$
B
$$\left( { - \infty ,0} \right)$$
C
$$\left( { - \infty ,\infty } \right) - \left\{ 0 \right\}$$
D
$$\left( { - \infty ,\infty } \right)$$
4
AIEEE 2009
+4
-1
Let $$f\left( x \right) = {\left( {x + 1} \right)^2} - 1,x \ge - 1$$

Statement - 1 : The set $$\left\{ {x:f\left( x \right) = {f^{ - 1}}\left( x \right)} \right\} = \left\{ {0, - 1} \right\}$$.

Statement - 2 : $$f$$ is a bijection.
A
Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1
B
Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1
C
Statement - 1 is true, Statement - 2 is false
D
Statement - 1 is false, Statement - 2 is true
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